calculating total current help

Discussion in 'The Projects Forum' started by Gadersd, Jan 1, 2013.

  1. Gadersd

    Thread Starter Member

    Dec 8, 2012
    Hi. I need help calculating the total current in this circuit with the transistor. How would I do it if the vbe is 0.7 v and the vce is 0.05 v?
  2. Gadersd

    Thread Starter Member

    Dec 8, 2012
    Oh and the voltage is 5 v.
  3. Jony130

    AAC Fanatic!

    Feb 17, 2009
    Simply find Ie current

    Ie = Ib + Ic

    Ib = (Vcc - Vbe)/(Rb + (Hfe+1)*Re) = (5V - 0.7V)/(10K + 101*10K)) = 4.3V/(10K + 1.01M) = 4.3V/1.02MΩ = 4.21μA

    Ic = Hfe*Ib = 421.56μA

    Ie = (Hfe+1)*Ib = 425.78μA

    Vce = Vcc - Ic*Rc - Ie*Re = 5V - 0.42156V - 4.2578V = 0.3V

    So BJT is in saturation region.

    Itot ≈ (Vcc - Vce)/(Rc+Re) ≈ 450μA
  4. WBahn


    Mar 31, 2012
    Where do you get that Hfe is 100? That wasn't given. Also, if the BJT is in saturation, then you know that Ic does NOT equal Hfe*Ib, since that is only valid in the active region. Finally, the problem specifically stated that Vce is 0.05V, not 0.3V.

    Let's first assume that the beta is infinite and that there is no base current. This means that Ve = Vcc - Vbe = 4.3V yielding an emitter current of 430μA. This is a maximum. Similarly, it is a maximum collector current, which results in a collector voltage of Vcc-Ic*Rc which yields 4.57V. This means that the minimum voltage that Vce can be is 0.27V. Thus the stated value of 0.05V is not possible.

    But let's ignore this for now, or assume that we are given valid values for Vbe and Vce. Under those constraints, there is no need to assume or know what the transistor β happens to be. In fact, we don't need to know that it is a transistor at all but instead can treat it as a black box with an unknown circuit having three terminals, C, B, and E.

    We have

    Ve = Ie*Re
    Vb = Ve + Vbe = Vcc - Ib*Rb
    Vc = Ve + Vce = Vcc - Ic*Rc
    Ie = Ib + Ic

    A bit of manipulation, and we have

    Ie*Re + Vbe = Vcc - Ib*Rb
    Ie*Re + Vce = Vcc - (Ie-Ib)*Rc


    Ie*Re + Ib*Rb = Vcc - Vbe
    Ie*(Re+Rc) - Ib*Rc = Vcc - Vce

    Note that the second equation above reduces to Jony130's final approximation under the assumption that the base current is negligible or, equivalently, that the beta is sufficiently high.

    If we aren't willing to make that assumption, then we can just proceed to eliminate it from the pair of equations

    Ie*(Re/Rb) + Ib = (Vcc - Vbe)/Rb
    Ie*(Re+Rc)/Rc - Ib = (Vcc - Vce)/Rc


    Ie*(Re/Rb + Re/Rc + 1) = Vcc(1/Rb + 1/Rc) - (Vbe/Rb + Vce/Rc)

    Ie*[1 + Re/(Rb||Rc)] = Vcc/(Rb||Rc) - (Vbe/Rb + Vce/Rc)


    Ie*[Re + (Rb||Rc)] = Vcc - (Vbe/Rb + Vce/Rc)*(Rb||Rc)

    Either way,

    Ie = Vcc/[Re + (Rb||Rc)] - (Vbe/Rb + Vce/Rc)/[1 + Re/(Rb||Rc)]

    If you are only looking for total current, here it is, except that you have to check the validity of the initial assumption that the transistor is on (either active or saturated). If we plug in the given values into this equation, we get an emitter current of 448μA, which seems quite reasonable. But when we ask if it is consistent with a BJT that is conducting, we discover that the base current would have to be -18μA (because the base voltage would be 5.18V.

    Thus the given values are not consistent with the circuit operating. Either the Vce is larger or the Vbe is smaller.

    A useful exercise might be to map out the Vce vs. Vbe for which this circuit uields valid results assuming infinite beta. An even more useful exercise might be to extend this graph by plotting similar lines for different values of beta.
  5. Gadersd

    Thread Starter Member

    Dec 8, 2012
    That seems like a lot of work just to find the total current. Are there any online programs that will find it for me?
  6. Wendy


    Mar 24, 2008
    Heh, it is much better to learn this math, this is where understanding comes from. Are you a hobbiest or a student? If you plan on a long term relationship with electronics you will find many other examples where you need the math, not all circuits have an online calculator.

    It is only simple algebra, after a while you can do most of the steps in your head. It is the concepts you truly need to get down, such as Ic ≈ Ie (if beta β were infinite they would equal exactly).

    As Bahn mentioned, beta with transistors is variable (depending on saturation, among other things), which screws things up nicely. Fortunately, when in linear mode β is mostly fixed. Another rule of thumb, when the transistor is in saturation (that is, Vce is less than 0.2V) it takes at least Ib/10 = Ic .

    Way back when (a couple of years out of college) I decided to really learn the math. I made a program to calculate the basic transistor configurations, which I then almost never used. It was on a TRS-80, and is long gone now. Transistor math I have down though.
    Last edited: Jan 1, 2013
  7. #12


    Nov 30, 2010
    Sure. Just google, "finding current through transistors that can't exist".

    Are you joking? Even if there is a calculator somewhere on the internet, what do you expect it to do with a circuit that isn't possible?

    Edit: Bill must be feeling a lot nicer than I feel today. The "point and click" generation kind of makes my head want to explode.
  8. Gadersd

    Thread Starter Member

    Dec 8, 2012
    One more thing. How would I calculate the Re with a led in it?
  9. WBahn


    Mar 31, 2012
    If you put an LED in it (I assume by "in it" you are referring to putting an LED in series with Re) then you have a different circuit and you analyze the new circuit. Walk the voltages and currents around the circuits using KVL and KCL making whatever simplifying assumptions you believe are reasonable.

    Part of the problem is deciding what simplifications are reasonable and this is contingent upon understanding exactly what it is you are trying to find and part of this is an awareness of how accurate/precise/exact the answer that you are looking for needs to be.